# How do you find the product of (t+1)(t^2+2t+4)?

Aug 13, 2017

$\left(t + 1\right) \left({t}^{2} + 2 t + 4\right) = {t}^{3} + 3 {t}^{2} + 6 t + 4$

#### Explanation:

You can use the distributive property to find:

$\left(t + 1\right) \left({t}^{2} + 2 t + 4\right) = t \left({t}^{2} + 2 t + 4\right) + 1 \left({t}^{2} + 2 t + 4\right)$

$\textcolor{w h i t e}{\left(t + 1\right) \left({t}^{2} + 2 t + 4\right)} = \left({t}^{3} + 2 {t}^{2} + 4 t\right) + \left({t}^{2} + 2 t + 4\right)$

$\textcolor{w h i t e}{\left(t + 1\right) \left({t}^{2} + 2 t + 4\right)} = {t}^{3} + 2 {t}^{2} + {t}^{2} + 4 t + 2 t + 4$

$\textcolor{w h i t e}{\left(t + 1\right) \left({t}^{2} + 2 t + 4\right)} = {t}^{3} + 3 {t}^{2} + 6 t + 4$

Alternatively, just focus on each power of $t$ in descending order and add together the different ways of getting it by multiplying a term in the first factor by a term in the second...

${t}^{3} : \text{ } t \cdot {t}^{2} = {t}^{3}$

${t}^{2} : \text{ } t \left(2 t\right) + 1 \left({t}^{2}\right) = 3 {t}^{2}$

$t : \text{ } t \left(4\right) + 1 \left(2 t\right) = 6 t$

$1 : \text{ } 1 \left(4\right) = 4$

$\left(t + 1\right) \left({t}^{2} + 2 t + 4\right) = {t}^{3} + 3 {t}^{2} + 6 t + 4$

When looking at each power of $t$, we only really need to multiply the coefficients and add the products, so:

${t}^{3} : \text{ } 1 \cdot 1 = 1$

${t}^{2} : \text{ } 1 \cdot 2 + 1 \cdot 1 = 3$

$t : \text{ } 1 \cdot 4 + 1 \cdot 2 = 6$

$1 : \text{ } 1 \cdot 4 = 4$

With practice you will probably find that you can normally do these calculations in your head as you write down the product:

$\left(t + 1\right) \left({t}^{2} + 2 t + 4\right) = {t}^{3} + 3 {t}^{2} + 6 t + 4$